Let's solve the problem step by step: --- ### Step 1: **Understand the Region Q** Given: - **Sphere**: \( z = \sqrt{36-x^2-y^2} \implies x^2 + y^2 + z^2 = 36 \) - **Cone**: \( z = \sqrt{x^2 + y^2} \) - **Plane**: \( x \geq 0 \) The region \( Q \) is the volume inside the sphere, above the cone, and on the right of the plane \( x = 0 \) (i.e., in the half-space where \( x \geq 0 \)). --- ### Step 2: **Draw the Diagram** (for a & b) You can use a tool like GeoGebra, but here's a description for manual sketch: #### a) **3D Diagram** - Draw a sphere of radius 6 centered at the origin. - Draw a cone opening upwards from the origin (with half-angle \( \theta = \arctan(1) = 45^\circ \)). - Indicate the half-space where \( x \geq 0 \) (the right half of the 3D space). #### b) **Projection onto the xy-plane** - The intersection of the cone and the sphere gives \( x^2 + y^2 + (\sqrt{x^2 + y^2})^2 = 36 \implies x^2 + y^2 + x^2 + y^2 = 36 \implies 2(x^2 + y^2) = 36 \implies x^2 + y^2 = 18 \). - In the \( x \geq 0 \) half-plane, the projection is the right half of the circle \( x^2 + y^2 = 18 \). --- ### Step 3: **Set Up the Integral in Spherical Coordinates** #### **Coordinate Transformations** - \( x = r \sin\theta \cos\phi \) - \( y = r \sin\theta \sin\phi \) - \( z = r \cos\theta \) - Volume element: \( dV = r^2 \sin\theta \, dr \, d\theta \, d\phi \) #### **Region Bounds** - **Sphere**: \( r = 0 \) to \( r = 6 \) - **Cone**: \( z = \sqrt{x^2 + y^2} \implies r \cos\theta = r \sin\theta \implies \theta = \frac{\pi}{4} \) - So, \( \theta \) from 0 (z-axis) to \( \frac{\pi}{4} \) (cone) - **Plane \( x \geq 0 \)**: \( x = r \sin\theta \cos\phi \geq 0 \implies \cos\phi \geq 0 \implies \phi \in [-\frac{\pi}{2}, \frac{\pi}{2}] \) #### **Integral to Evaluate** Given: \( x^2(x^2 + y^2 + z^2) \) - \( x^2 + y^2 + z^2 = r^2 \) - \( x^2 = r^2 \sin^2\theta \cos^2\phi \) So, \[ x^2(x^2 + y^2 + z^2) = r^2 \sin^2\theta \cos^2\phi \cdot r^2 = r^4 \sin^2\theta \cos^2\phi \] --- ### Step 4: **Write the Triple Integral** \[ \int_{\phi = -\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{\theta = 0}^{\frac{\pi}{4}} \int_{r = 0}^{6} \left[ r^4 \sin^2\theta \cos^2\phi \right] \cdot r^2 \sin\theta \, dr \, d\theta \, d\phi \] Simplify the integrand: - \( r^4 \cdot r^2 = r^6 \) - \( \sin^2\theta \cdot \sin\theta = \sin^3\theta \) So, \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{0}^{6} r^6 \sin^3\theta \cos^2\phi \, dr \, d\theta \, d\phi \] --- ### Step 5: **Evaluate Each Integral** #### **1. Integrate with respect to \( r \):** \[ \int_0^6 r^6 \, dr = \left[ \frac{r^7}{7} \right]_0^6 = \frac{6^7}{7} = \frac{279936}{7} \] --- #### **2. Integrate with respect to \( \theta \):** \[ \int_0^{\frac{\pi}{4}} \sin^3\theta \, d\theta \] Recall: \[ \sin^3\theta = \sin\theta (1 - \cos^2\theta) = \sin\theta - \sin\theta \cos^2\theta \] Let’s break it up: \[ \int \sin^3\theta d\theta = \int \sin\theta d\theta - \int \sin\theta \cos^2\theta d\theta \] - \( \int \sin\theta d\theta = -\cos\theta \) - For \( \int \sin\theta \cos^2\theta d\theta \), set \( u = \cos\theta \), \( du = -\sin\theta d\theta \): \[ \int \sin\theta \cos^2\theta d\theta = -\int u^2 du = -\frac{u^3}{3} = -\frac{\cos^3\theta}{3} \] So, \[ \int \sin^3\theta d\theta = -\cos\theta + \frac{\cos^3\theta}{3} \] Evaluate from \( 0 \) to \( \frac{\pi}{4} \): - \( \cos 0 = 1 \), \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) - \( (\frac{\sqrt{2}}{2})^3 = \frac{2^{3/2}}{8} = \frac{\sqrt{2}}{4} \) So, \[ = \left[ -\cos\theta + \frac{\cos^3\theta}{3} \right]_0^{\frac{\pi}{4}} = \left( -\frac{\sqrt{2}}{2} + \frac{1}{3} \frac{\sqrt{2}}{4} \right) - \left( -1 + \frac{1}{3}\cdot 1 \right) \] \[ = \left(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{12}\right) - \left(-1 + \frac{1}{3}\right) \] \[ = \left(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{12}\right) + 1 - \frac{1}{3} \] \[ = -\frac{6\sqrt{2}}{12} + \frac{\sqrt{2}}{12} + \frac{2}{3} \] \[ = -\frac{5\sqrt{2}}{12} + \frac{2}{3} \] --- #### **3. Integrate with respect to \( \phi \):** \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2\phi \, d\phi \] Recall: \[ \cos^2\phi = \frac{1 + \cos(2\phi)}{2} \] \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2\phi d\phi = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, d\phi + \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 2\phi \, d\phi \] - The second term is zero because \( \cos 2\phi \) is symmetric and integrates to zero over the interval. - The first term: \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, d\phi = \pi \) So, \[ = \frac{1}{2} \pi \] --- ### Step 6: **Combine All Results** Multiply all results: \[ \boxed{ \frac{279936}{7} \times \left( -\frac{5\sqrt{2}}{12} + \frac{2}{3} \right) \times \frac{\pi}{2} } \] --- #### **Final Simplified Answer:** \[ \boxed{ \int \int \int_Q x^2 (x^2 + y^2 + z^2) dV = \frac{279936}{7} \cdot \left(\frac{2}{3} - \frac{5\sqrt{2}}{12} \right) \cdot \frac{\pi}{2} } \] Or, as a single fraction: \[ = \frac{279936 \pi}{14} \left( \frac{2}{3} - \frac{5\sqrt{2}}{12} \right) \] --- #### **Decimal Approximation (optional):** - \( 279936/7 \approx 39990.857 \) - \( \frac{2}{3} - \frac{5\sqrt{2}}{12} \approx 0.6667 - 0.5893 = 0.0774 \) - \( \frac{\pi}{2} \approx 1.5708 \) So, the value is approximately: \[ 39990.857 \times 0.0774 \times 1.5708 \approx 4858.5 \] --- ## **Summary** **(c) The required triple integral is:** \[ \boxed{ \frac{279936}{7} \cdot \left(\frac{2}{3} - \frac{5\sqrt{2}}{12} \right) \cdot \frac{\pi}{2} } \] --- **Let me know if you need the diagrams for (a) and (b) in GeoGebra or a hand-drawn version!**

Certainly! Let's proceed step-by-step, with detailed explanations, calculations, and guidance for diagrams. --- # Problem Recap: **Region \(Q\)** is bounded by: - The sphere: \(x^2 + y^2 + z^2 = 36\) - The cone: \(z = \sqrt{x^2 + y^2}\) - The plane: \(x \geq 0\) We need to evaluate the integral: \[ I = \iiint_Q x^2 (x^2 + y^2 + z^2) \, dV \] --- # Step 1: Understand the region \(Q\) - The sphere: radius \(6\), centered at the origin. - The cone: \(z = \sqrt{x^2 + y^2}\), which is a double-napped cone opening upwards with a 45° half-angle (\(\tan \theta = 1\)). - The plane: \(x \geq 0\), the right half-space. **Visualizing the region:** - Inside the sphere, above the cone, and on the side \(x \geq 0\). --- # Step 2: Drawing diagrams (a) and (b) While I cannot generate images directly here, I will describe how you can draw them: ### (a) 3D Diagram: - Draw a sphere with radius 6 centered at the origin. - Inside it, sketch a cone originating at the origin, opening upward with a 45° angle. - Highlight the half-space where \(x \geq 0\). The region \(Q\) is the part of the sphere above the cone and in the half-space \(x \geq 0\). ### (b) Projection onto the xy-plane: - Find the intersection of the cone and sphere: \[ z = \sqrt{x^2 + y^2} \] and \[ x^2 + y^2 + z^2 = 36 \] Substitute \(z\): \[ x^2 + y^2 + x^2 + y^2 = 36 \Rightarrow 2(x^2 + y^2) = 36 \Rightarrow x^2 + y^2 = 18 \] - Since \(x \geq 0\), the projection is the right half of the circle \(x^2 + y^2 = 18\). --- # Step 3: Choosing Coordinates Given the symmetry and bounds, **spherical coordinates** are suitable: \[ \begin{cases} x = r \sin \theta \cos \phi \\ y = r \sin \theta \sin \phi \\ z = r \cos \theta \end{cases} \] with the Jacobian: \[ dV = r^2 \sin \theta \, dr \, d\theta \, d\phi \] --- # Step 4: Determine bounds in spherical coordinates ### 1. Radius \(r\): - Inside the sphere: \(0 \leq r \leq 6\) ### 2. \(\theta\): - Cone: \(z = r \cos \theta = r \sin \theta\) - So, \(\cos \theta = \sin \theta \Rightarrow \theta = \pi / 4\) - Region above the cone: \(0 \leq \theta \leq \pi/4\) ### 3. \(\phi\): - Plane \(x \geq 0\): \[ x = r \sin \theta \cos \phi \geq 0 \Rightarrow \cos \phi \geq 0 \Rightarrow -\pi/2 \leq \phi \leq \pi/2 \] --- # Step 5: Express the integrand in spherical coordinates \[ x^2(x^2 + y^2 + z^2) = x^2 \cdot r^2 \] since \(x^2 + y^2 + z^2 = r^2\). - \(x = r \sin \theta \cos \phi\), so \[ x^2 = r^2 \sin^2 \theta \cos^2 \phi \] Thus, the integrand: \[ x^2 (x^2 + y^2 + z^2) = r^2 \sin^2 \theta \cos^2 \phi \cdot r^2 = r^4 \sin^2 \theta \cos^2 \phi \] --- # Step 6: Set up the triple integral \[ I = \int_{\phi = -\pi/2}^{\pi/2} \int_{\theta=0}^{\pi/4} \int_{r=0}^{6} r^4 \sin^2 \theta \cos^2 \phi \times r^2 \sin \theta \, dr \, d\theta \, d\phi \] Simplify: \[ I = \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi/4} \int_{0}^{6} r^{6} \sin^{3} \theta \cos^{2} \phi \, dr \, d\theta \, d\phi \] --- # Step 7: Integrate with respect to \(r\) \[ \int_0^{6} r^{6} dr = \left[\frac{r^{7}}{7}\right]_0^{6} = \frac{6^{7}}{7} = \frac{279936}{7} \] --- # Step 8: Integrate with respect to \(\theta\): \[ J_\theta = \int_0^{\pi/4} \sin^{3} \theta \, d\theta \] **Calculating \(\int \sin^3 \theta \, d\theta\):** Use the identity: \[ \sin^{3} \theta = \sin \theta (1 - \cos^2 \theta) \] Substitute \(u = \cos \theta\), \(du = -\sin \theta d\theta\): \[ J_\theta = \int \sin \theta (1 - \cos^2 \theta) d\theta \] \[ = -\int (1 - u^2) du \] \[ = -\left[ u - \frac{u^3}{3} \right] + C \] Back to \(\theta\): \[ J_\theta = -\left[\cos \theta - \frac{\cos^{3} \theta}{3} \right] \] Evaluate from \(0\) to \(\pi/4\): - \(\cos 0 = 1\) - \(\cos (\pi/4) = \frac{\sqrt{2}}{2}\) \[ J_\theta = - \left[ \frac{\sqrt{2}}{2} - \frac{(\sqrt{2}/2)^3}{3} \right] + \left[ 1 - \frac{1}{3} \right] \] Calculate \((\sqrt{2}/2)^3\): \[ (\sqrt{2}/2)^3 = \frac{\sqrt{2}^3}{2^3} = \frac{2^{3/2}}{8} = \frac{2 \sqrt{2}}{8} = \frac{\sqrt{2}}{4} \] Now, \[ J_\theta = - \left[ \frac{\sqrt{2}}{2} - \frac{\sqrt{2}/4}{3} \right] + \left( \frac{2}{3} \right) \] \[ J_\theta = - \left[ \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{12} \right] + \frac{2}{3} \] Express numerator with common denominators: \[ \frac{\sqrt{2}}{2} = \frac{6 \sqrt{2}}{12}, \quad \frac{\sqrt{2}}{12} \text{ stays the same} \] So, \[ J_\theta = - \left[ \frac{6 \sqrt{2}}{12} - \frac{\sqrt{2}}{12} \right] + \frac{2}{3} = - \left( \frac{5 \sqrt{2}}{12} \right) + \frac{2}{3} \] --- # Step 9: Integrate with respect to \(\phi\): \[ J_\phi = \int_{-\pi/2}^{\pi/2} \cos^2 \phi \, d\phi \] Recall: \[ \cos^2 \phi = \frac{1 + \cos 2\phi}{2} \] Thus, \[ J_\phi = \frac{1}{2} \int_{-\pi/2}^{\pi/2} 1 \, d\phi + \frac{1}{2} \int_{-\pi/2}^{\pi/2} \cos 2\phi \, d\phi \] - \(\int_{-\pi/2}^{\pi/2} 1 \, d\phi = \pi\) - \(\int_{-\pi/2}^{\pi/2} \cos 2\phi \, d\phi = 0\) (since \(\cos 2\phi\) is symmetric and integrates to zero over symmetric bounds) Therefore, \[ J_\phi = \frac{1}{2} \times \pi = \frac{\pi}{2} \] --- # **Final Calculation:** Putting all parts together: \[ I = \left( \frac{279936}{7} \right) \times \left( - \frac{5 \sqrt{2}}{12} + \frac{2}{3} \right) \times \frac{\pi}{2} \] --- # **Final Answer:** \[ \boxed{ I = \frac{279936}{7} \times \left( \frac{2}{3} - \frac{5 \sqrt{2}}{12} \right) \times \frac{\pi}{2} } \] --- # **Optional Numerical Approximation:** - \(279936/7 \approx 39990.857\) - \(\frac{2}{3} \approx 0.6667\) - \(\frac{5 \sqrt{2}}{12} \approx 0.5893\) Compute the difference: \[ 0.6667 - 0.5893 = 0.0774 \] Now multiply: \[ 39990.857 \times 0.0774 \times 1.5708 \approx 4858.5 \] **Thus, the approximate value of the integral is about \(\boxed{4858.5}\).** --- # **Summary:** - **Diagrams:** - Draw the sphere, cone, and half-space \(x \geq 0\) for the 3D diagram. - Project the intersection on xy-plane as the right half of the circle \(x^2 + y^2 = 18\). - **Integral setup:** \[ I = \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi/4} \int_{0}^{6} r^6 \sin^3 \theta \cos^2 \phi \, dr \, d\theta \, d\phi \] - **Final result:** \[ \boxed{ I = \frac{279936}{7} \left( \frac{2}{3} - \frac{5 \sqrt{2}}{12} \right) \frac{\pi}{2} } \] Please let me know if you'd like me to prepare diagrams in a specific software or further clarify any step!

Final Answers: **(a) and (b) Diagrams:** - *Description:* The region is bounded inside a sphere of radius 6, above the cone \(z = \sqrt{x^2 + y^2}\) (a 45° cone), and in the half-space \(x \geq 0\). The projection onto the xy-plane is the right half of the circle \(x^2 + y^2 = 18\). --- **(c) Value of the integral:** \[ \boxed{ I = \frac{279936}{7} \times \left( \frac{2}{3} - \frac{5 \sqrt{2}}{12} \right) \times \frac{\pi}{2} } \] **Numerical approximation:** \[ I \approx 4858.5 \]